Q.How will you bring about the following conversions in not more than two steps?
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Oxidation Reactions
Alcohol Oxidation: The Intuition First
Imagine you have a molecule of ethanol — the alcohol in your hand sanitizer or a drink. It has a carbon atom bonded to an –OH group. Now picture that –OH group as a "handle" that can be transformed. Oxidation, in organic chemistry, doesn't always mean adding oxygen — it often means removing hydrogen from a carbon that already has a bond to oxygen. For alcohols, oxidation is like "stripping away" hydrogen atoms from the carbon that holds the –OH, turning the alcohol into a more oxidized functional group.
Think of it this way: a primary alcohol (R–CH₂–OH) has two hydrogens on the carbon with the –OH. If you remove one hydrogen and the hydrogen from the –OH, you get an aldehyde (R–CHO). Remove both hydrogens (and the –OH hydrogen), and you get a carboxylic acid (R–COOH). A secondary alcohol (R–CHOH–R') has only one hydrogen on that carbon — remove it, and you get a ketone (R–CO–R'). A tertiary alcohol has no hydrogen on that carbon — so it cannot be oxidized without breaking the carbon skeleton.
That's the core intuition: oxidation of an alcohol is about removing hydrogens from the carbon bearing the –OH group. The more hydrogens you can remove, the more oxidized the product.
The Precise Statement
Alcohol oxidation is the process in which an alcohol loses hydrogen atoms (dehydrogenation) from the carbon bonded to the –OH group, increasing the number of C–O bonds (or decreasing C–H bonds). The outcome depends on the class of the alcohol:
| Alcohol Class | Structure | Product after oxidation | Reagent example |
|---|---|---|---|
| Primary (1°) | R–CH₂–OH | Aldehyde (R–CHO) then Carboxylic acid (R–COOH) | PCC (stops at aldehyde); K₂Cr₂O₇/H⁺ (goes to acid) |
| Secondary (2°) | R–CHOH–R' | Ketone (R–CO–R') | K₂Cr₂O₇/H⁺, CrO₃, etc. |
| Tertiary (3°) | R₃C–OH | No reaction (under normal conditions) | — |
A common mistake: students think "oxidation" always adds oxygen. For alcohols, it's removal of hydrogen from the carbon with the –OH. The oxygen from the –OH stays — it's the hydrogens that leave.
Why Does Tertiary Alcohol Not Oxidize?
Look at the carbon with the –OH in a tertiary alcohol: it has three carbon groups attached and no hydrogen. To form a C=O bond, you'd need to remove a hydrogen from that carbon — but there is none. The only way to oxidize a tertiary alcohol is to break a C–C bond (strong and difficult), which is not typical oxidation. So in standard organic chemistry, tertiary alcohols are inert to mild oxidizing agents.
A Real-World Analogy
Think of the alcohol carbon as a "parking spot" with a certain number of hydrogen "cars." Primary alcohol has two cars parked. Oxidation is like towing away one car (→ aldehyde) or both cars (→ carboxylic acid). Secondary alcohol has one car — tow it away, and you get a ketone. Tertiary alcohol has zero cars — nothing to tow, so no reaction.
Key Reagents to Remember (for exams)
- PCC (pyridinium chlorochromate): oxidizes 1° alcohols to aldehydes only — stops there.
- K₂Cr₂O₇ / H₂SO₄ (Jones reagent): oxidizes 1° alcohols all the way to carboxylic acids; 2° alcohols to ketones.
- KMnO₄: similar to dichromate, but stronger — can over-oxidize. …
Why this formula?
Oxidation Reactions: Why the Key Principles Hold
Oxidation reactions are fundamental to chemistry, and understanding why they work the way they do is essential for mastering Indian board exams (Class 11, 12, JEE, NEET). Let's break down the core ideas from first principles.
1. The Core Definition: What Does "Oxidation" Really Mean?
Historically, oxidation meant "adding oxygen." But that's too narrow. The modern, exam-correct definition is:
Oxidation is the loss of electrons by a species.
This is the electronic concept (given by the ionic theory). The why behind this definition comes from the behavior of atoms during chemical reactions.
Why do atoms lose electrons?
Atoms seek stability. They achieve this by having a full outer electron shell (octet, duplet, or pseudo-inert gas configuration).
- Metals (like Na, Mg, Fe) have few valence electrons (1, 2, or 3). It's energetically easier for them to lose these electrons than to gain 5, 6, or 7.
- Non-metals (like O, Cl, F) have many valence electrons (5, 6, or 7). It's energetically easier for them to gain electrons.
So, when a metal reacts with a non-metal, the metal loses electrons (gets oxidized), and the non-metal gains electrons (gets reduced).
Example:
2Na+Cl2→2NaCl
- Na loses 1 electron: Na→Na++e− (Oxidation)
- Cl gains 1 electron: Cl2+2e−→2Cl− (Reduction)
Key takeaway: Oxidation and reduction always happen together (Redox reactions). You cannot have one without the other.
2. The Key Formula(e): Oxidation Number Rules
The oxidation number (O.N.) is a bookkeeping tool. It's not a real charge (except in ionic compounds), but it helps track electron flow.
Why do we assign oxidation numbers?
Because in covalent compounds (like CH4 or H2O), electrons are shared, not transferred. We need a way to pretend they are transferred to see which atom "owns" the electrons more.
The Rules (and why they exist)
| Rule | Statement | Why this rule? |
|---|---|---|
| 1 | O.N. of an element in its free state = 0 | No electron transfer has occurred. |
| 2 | O.N. of a monatomic ion = its charge | The atom has actually lost/gained that many electrons. |
| 3 | O.N. of H = +1 (except in metal hydrides where it's -1) | H is less electronegative than O, F, Cl, but more electronegative than metals. |
| 4 | O.N. of O = -2 (except in peroxides where it's -1, superoxides -1/2, and with F where it's +2) | O is highly electronegative (3.44 on Pauling scale). It "pulls" electrons toward itself. |
| 5 | Sum of O.N. in a neutral compound = 0 | The compound has no net charge. |
| 6 | Sum of O.N. in a polyatomic ion = charge of the ion | The ion's overall charge must be accounted for. |
The Derivation of a Key Formula: Finding O.N. of an Unknown Element
Suppose you need to find the O.N. of S in H2SO4.
Step 1: Write known O.N.s:
- H: +1 (rule 3)
- O: -2 (rule 4)
- S: let it be x (unknown)
Step 2: Apply rule 5 (neutral compound sum = 0):
2(+1)+x+4(−2)=0
Step 3: Solve:
2+x−8=0
x−6=0
x=+6
Why this works: The oxidation number is a mathematical consequence of the electronegativity hierarchy. Oxygen is more electronegative than sulfur, so it "takes" the electrons. Hydrogen is less electronegative than sulfur, so it "gives" electrons to sulfur. The net result is that sulfur appears to have lost 6 electrons.
3. The Key Formula(e): Balancing Redox Equations
Two methods are exam-critical: Oxidation Number Method and Ion-Electron Method (Half-Reaction Method).
Why do we need these methods?
Because in a redox reaction, the total number of electrons lost (oxidation) must equal the total number of electrons gained (reduction). This is the Law of Conservation of Charge.
The Ion-Electron Method (for acidic medium) — Step-by-step why
Example: Balance MnO4−+Fe2+→Mn2++Fe3+ (acidic)
Step 1: Write half-reactions.
-
Oxidation: Fe2+→Fe3++e−
Why? Fe loses 1 electron (O.N. goes from +2 to +3).
-
Reduction: MnO4−→Mn2+
Why? Mn gains electrons (O.N. goes from +7 to +2).
Step 2: Balance atoms other than H and O.
- Mn is already balanced (1 on each side).
Step 3: Balance O by adding H2O.
- Left: 4 O atoms. Right: 0 O atoms.
- Add 4 H2O to the right:
MnO4−→Mn2++4H2O
Step 4: Balance H by adding H+ (because acidic medium).
- Right: 8 H atoms (from 4 H2O). Left: 0 H atoms.
- Add 8 H+ to the left:
8H++MnO4−→Mn2++4H2O
Step 5: Balance charge by adding electrons.
- Left: 8(+1)+(−1)=+7 charge.
- Right: +2 charge.
- To make left = right, add 5 electrons to the left:
8H++MnO4−+5e−→Mn2++4H2O
Step 6: Multiply half-reactions to equalize electrons.
- Oxidation: Fe2+→Fe3++e− (×5)
- Reduction: 8H++MnO4−+5e−→Mn2++4H2O (×1)
Step 7: Add them:
5Fe2++8H++MnO4−→5Fe3++Mn2++4H2O
Why this works: Every step is driven by conservation laws:
- Mass balance: Same number of each atom on both sides.
- Charge balance: Net charge on left = net charge on right.
- Electron balance: Electrons lost = electrons gained.
4. The Key Formula(e): Electrochemical Series and Cell Potential
For a galvanic cell (voltaic cell), the cell potential Ecell∘ is:
Ecell∘=Ecathode∘−Eanode∘
Why this formula? …
(i) Propanone to Propene
Concept: Reduction followed by dehydration.
Steps:
- Reduce propanone with NaBH4 or LiAlH4 to propan-2-ol. …
Each conversion is achieved in two steps using standard organic reactions: reduction, oxidation, Grignard, Friedel-Crafts, aldol/hydrogenation, or nitration. The key is to choose the correct reagent sequence that gives the target in minimal steps, and to pick reagents that are actually chemoselective for the bond being changed.
(i) Propanone to Propene
- Reduce propanone to propan-2-ol using NaBH4 or LiAlH4. CH3COCH3 --NaBH4/H2O--> CH3CH(OH)CH3
- Dehydrate the alcohol to propene using conc. H2SO4 at 443 K (acid-catalysed E1 elimination). …
Here is a clear, concept-first solution for each conversion, using named reactions and stepwise logic.
(i) Propanone to Propene
Method: Reduction followed by Dehydration (2 steps)
- Step 1 (Reduction): Reduce propanone (CH3COCH3) to propan-2-ol using NaBH4 (or LiAlH4 / catalytic hydrogenation).
CH3COCH3NaBH4CH3CH(OH)CH3
- Step 2 (Dehydration): Heat propan-2-ol with concentrated H2SO4 at 170°C to eliminate water, forming propene.
CH3CH(OH)CH3conc. H2SO4ΔCH3CH=CH2
Key concept: Ketone → Alcohol → Alkene (via acid-catalyzed elimination).
(ii) Benzoic acid to Benzaldehyde
Method: Rosenmund Reduction (1 step)
- Step 1: Convert benzoic acid to benzoyl chloride using SOCl2 (or PCl5).
C6H5COOHSOCl2C6H5COCl
- Step 2: Reduce benzoyl chloride to benzaldehyde using H2 / Pd-BaSO4 (poisoned catalyst, Rosenmund reduction).
C6H5COClH2, Pd-BaSO4C6H5CHO
Key concept: Acid chloride → Aldehyde (selective hydrogenation stops at aldehyde).
(iii) Ethanol to 3-Hydroxybutanal
Method: Aldol Condensation (1 step)
- Step 1: Oxidize ethanol to ethanal using acidified K2Cr2O7 (or PCC).
CH3CH2OHK2Cr2O7/H+CH3CHO
- Step 2: Perform aldol condensation of ethanal with dilute NaOH at room temperature.
2 CH3CHOdil. NaOHCH3CH(OH)CH2CHO
Key concept: Two molecules of aldehyde combine to form a β-hydroxy aldehyde.
(iv) Benzene to m-Nitroacetophenone
Method: Friedel-Crafts Acylation followed by Nitration (2 steps)
- Step 1 (Acylation): Treat benzene with acetyl chloride (CH3COCl) and anhydrous AlCl3 to form acetophenone.
C6H6CH3COCl, AlCl3C6H5COCH3
- Step 2 (Nitration): Treat acetophenone with a nitrating mixture (conc. HNO3 + conc. H2SO4). The –COCH3 group is meta-directing, so the product is m-nitroacetophenone.
C6H5COCH3HNO3/H2SO4m-O2N-C6H4COCH3
Key concept: Meta-directing acyl group controls the position of the second substituent.
(v) Benzaldehyde to Benzophenone
Method: Grignard Reaction followed by Oxidation (2 steps)
- Step 1 (Grignard addition): React benzaldehyde with phenylmagnesium bromide (C6H5MgBr), then hydrolyze to form diphenylmethanol.
C6H5CHO1. C6H5MgBr2. H3O+C6H5CH(OH)C6H5
- Step 2 (Oxidation): Oxidize diphenylmethanol to benzophenone using PCC (or acidified K2Cr2O7).
C6H5CH(OH)C6H5PCCC6H5COC6H5
Key concept: Aldehyde + Grignard → Secondary alcohol → Ketone.
(vi) Bromobenzene to 1-Phenylethanol
Method: Grignard Reaction with Ethanal (2 steps)
- Step 1 (Grignard formation): Treat bromobenzene with Mg in dry ether to form phenylmagnesium bromide.
C6H5BrMg, dry etherC6H5MgBr
- Step 2 (Addition to aldehyde): React the Grignard reagent with ethanal (CH3CHO), then hydrolyze to get 1-phenylethanol.
C6H5MgBr1. CH3CHO2. H3O+C6H5CH(OH)CH3
Key concept: Aryl halide → Grignard → Secondary alcohol.
(vii) Benzaldehyde to 3-Phenylpropan-1-ol
Method: Aldol Condensation followed by Hydrogenation (2 steps)
- Step 1 (Crossed Aldol): React benzaldehyde with ethanal in dilute NaOH to form cinnamaldehyde (3-phenylprop-2-enal). C6H5CHO+CH3CHOdil. NaOHC6H5CH=CHCHO …
Common Mistakes in Oxidation Reaction Conversions (CBSE/JEE)
Students often lose marks in these conversions due to reagent confusion, overlooking reaction conditions, and ignoring functional group priorities. Below is a mistake-by-mistake breakdown.
(i) Propanone → Propene
Correct route:
Propanone NaBH4 Propan-2-ol Conc. H2SO4,Δ Propene
Common mistakes:
- ✗ Using KMnO₄ or K₂Cr₂O₇ (oxidising agents) — these will not reduce the ketone.
- ✗ Trying direct dehydration of propanone — ketones do not dehydrate like alcohols.
- ✗ Using LiAlH₄ instead of NaBH₄ — LiAlH₄ works but is less common in 2-step problems; NaBH₄ is safer and sufficient.
How to avoid:
Remember: Ketone → Alcohol (reduction) → Alkene (dehydration). Use NaBH₄ for reduction, then conc. H₂SO₄ with heat.
(ii) Benzoic acid → Benzaldehyde
Correct route:
Benzoic acid SOCl2 Benzoyl chloride H2,Pd-BaSO4(Rosenmund reduction) Benzaldehyde
Common mistakes:
- ✗ Using LiAlH₄ — this reduces benzoic acid all the way to benzyl alcohol, not benzaldehyde.
- ✗ Using NaBH₄ — NaBH₄ does not reduce carboxylic acids.
- ✗ Forgetting to convert to acid chloride first — direct reduction of –COOH to –CHO is not possible in one step.
How to avoid:
Always convert –COOH to –COCl first, then use Rosenmund reduction (H₂/Pd-BaSO₄). Never use strong reducing agents like LiAlH₄ here.
(iii) Ethanol → 3-Hydroxybutanal
Correct route:
Ethanol Cu, 573 K Ethanal Dil. NaOH, aldol condensation 3-Hydroxybutanal
Common mistakes:
- ✗ Using conc. NaOH — this leads to dehydration of the aldol product, giving crotonaldehyde instead.
- ✗ Trying to oxidise ethanol directly to 3-hydroxybutanal — impossible; you need two separate steps.
- ✗ Using KMnO₄ for oxidation — this overoxidises ethanol to acetic acid.
How to avoid:
First step: controlled oxidation (Cu/573 K or PCC) to get ethanal. Second step: aldol condensation with dilute NaOH at low temperature.
(iv) Benzene → m-Nitroacetophenone
Correct route:
Benzene CH3COCl, AlCl3 Acetophenone Conc. HNO3+Conc. H2SO4 m-Nitroacetophenone
Common mistakes:
- ✗ Nitrating before acylating — nitrobenzene is deactivated and cannot undergo Friedel-Crafts acylation.
- ✗ Using CH₃Cl + AlCl₃ (alkylation) — this gives toluene, then nitration gives o/p-nitrotoluene, not m-nitroacetophenone.
- ✗ Forgetting that –COCH₃ is meta-directing — students sometimes expect ortho/para products.
How to avoid:
Always do Friedel-Crafts acylation first, then nitration. The acyl group deactivates the ring and directs meta.
(v) Benzaldehyde → Benzophenone
Correct route:
Benzaldehyde \xrightarrow{\text{C}_6\text{H}_5\text{MgBr, \, \text{then H}_3\text{O}^+}} 1,2-Diphenylethanol PCC or CrO3 Benzophenone
Common mistakes:
- ✗ Using C₆H₅MgBr directly on benzaldehyde without protecting the aldehyde — Grignard reagent reacts with –CHO first.
- ✗ Trying Friedel-Crafts acylation with benzoyl chloride on benzene — that gives benzophenone, but starting from benzaldehyde requires a different route.
- ✗ Forgetting the oxidation step after Grignard addition.
How to avoid:
Add phenylmagnesium bromide to benzaldehyde (nucleophilic addition), then oxidise the secondary alcohol to ketone using PCC or CrO₃.
(vi) Bromobenzene → 1-Phenylethanol
Correct route:
Bromobenzene Mg, dry ether C₆H₅MgBr CH3CHO, then H3O+ 1-Phenylethanol
Common mistakes:
- ✗ Using CH₃COCl instead of CH₃CHO — this gives acetophenone, not the alcohol.
- ✗ Forgetting dry ether — Grignard formation fails in presence of moisture.
- ✗ Trying Wurtz-Fittig reaction — that gives alkylbenzene, not alcohol.
How to avoid:
Make Grignard reagent from bromobenzene, then react with ethanal (acetaldehyde), followed by hydrolysis.
(vii) Benzaldehyde → 3-Phenylpropan-1-ol
Correct route:
Benzaldehyde CH3CH2MgBr, then H3O+ 1-Phenylpropan-1-ol Red P + HI or H2/Pd 3-Phenylpropan-1-ol
Common mistakes:
- ✗ Using CH₃MgBr — this gives 1-phenylethanol (only 2 carbons added).
- ✗ Forgetting the reduction step — Grignard addition gives a secondary alcohol, not the primary alcohol required.
- ✗ Using LiAlH₄ on benzaldehyde directly — that gives benzyl alcohol, not 3-phenylpropan-1-ol.
How to avoid:
Use ethylmagnesium bromide (C₂H₅MgBr) to add two carbons, then reduce the –OH to –H using Red P/HI or catalytic hydrogenation.
(viii) Benzaldehyde → α-Hydroxyphenylacetic acid
Correct route: …
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